Here’s a quick proof. In fact, all we need to show is that the two second conditions are equivalent for III. First assume that for any a∈IaIa\in I and b∈LbLb\in L, a∧b∈IabIa\wedge b\in I. If b≤abab\leq a, then b=a∧b∈IbabIb=a\wedge b\in I. Conversely, since a∧b≤a∈IabaIa\wedge b\leq a\in I, a∧b∈IabIa\wedge b\in I as well.

Special Ideals. Let III be an ideal of a lattice LLL. Below are some common types of ideals found in lattice theory.

ideal generated by a set. Let XXX be a subset of a lattice LLL. Let SSS be the set of all ideals of LLL containing XXX. Since S≠∅SS\neq\varnothing (L∈SLSL\in S), the intersectionMMM of all elements in SSS, is also an ideal of LLL that contains XXX. MMM is called the ideal generated byXXX, written (X]X(X]. If XXX is a singleton{x}x\{x\}, then MMM is said to be a principal idealgenerated byxxx, written (x]x(x]. (Note that this construction can be easily carried over to the construction of a sublattice generated by a subset of a lattice).

Remarks. Let LLL be a lattice.

1.

Given any subset X⊂LXLX\subset L, let X′superscriptXnormal-′X^{{\prime}} be the set consisting of all finitejoins of elements of XXX, which is clearly a directed set. Then ↓X′normal-↓absentsuperscriptXnormal-′\downarrow\!\!X^{{\prime}}, the down set of X′superscriptXnormal-′X^{{\prime}}, is (X]X(X]. Any element of (X]X(X] is less than or equal to a finite join of elements of XXX.

2.

If LLL is a distributive lattice, every maximal ideal is prime. Suppose I⊆LILI\subseteq L is maximal and a∧b∈IabIa\wedge b\in I with a∉IaIa\notin I. Then the ideal generated by III and aaa must be LLL, so that b≤p∨abpab\leq p\vee a for some p∈IpIp\in I. Then b=b∧b≤(p∨a)∧b=(p∧b)∨(a∧b)∈IbbbpabpbabIb=b\wedge b\leq(p\vee a)\wedge b=(p\wedge b)\vee(a\wedge b)\in I, which meansb∈IbIb\in I. So III is prime.

3.

If LLL is a complemented lattice, every prime ideal is maximal. Suppose I⊆LILI\subseteq L is prime and a∉IaIa\notin I. Let bbb be a complement of aaa, then b∈IbIb\in I, for otherwise, 0=a∧b∉I0abI0=a\wedge b\notin I, a contradiction. Let JJJ be the ideal generated by III and aaa, then 1≤b∨a∈J1baJ1\leq b\vee a\in J, so J=LJLJ=L.

4.

Combining the two results above, in a Boolean algebra, an ideal is prime iff it is maximal.

Out of these, M,N,S,T,UMNSTUM,N,S,T,U are prime, and M,NMNM,N are maximal. The ideal generated by, say {c,e}ce\{c,e\}, is RRR. Looking more closely, we see that RRR can actually be generated by ccc, and so is principal. In fact, all ideals in LLL are principal, generated by their maximal elements. It is not hard to see, that in a lattice LLL with acc (ascending chain condition), all ideals are principal:

Proof.

. First, let’s show that an ideal III in a lattice LLL with acc has at least one maximal element. Suppose a∈IaIa\in I. If aaa is not maximal in III, there is a a1∈Isubscripta1Ia_{1}\in I such that a≤a1asubscripta1a\leq a_{1}. If a1subscripta1a_{1} is not maximal in III, repeat the process above so we get a chaina≤a1≤a2≤…asubscripta1subscripta2normal-…a\leq a_{1}\leq a_{2}\leq\ldots in III. Eventually this chain terminates an=an+1=⋯subscriptansubscriptan1normal-⋯a_{n}=a_{{n+1}}=\cdots. Thus b=anbsubscriptanb=a_{n} is maximal in III. Next, suppose that III has two distinct maximal elements. Then their join is again in III, contradicting maximality. So bbb is unique and all elements ccc such that c≤bcbc\leq b must be in III. Therefore, I=(b]fragmentsIfragmentsnormal-(bnormal-]I=(b].∎

Finally, an example of a sublattice that is not an ideal is the subset {b,c,d,e,0}bcde0\{b,c,d,e,0\}.